NCERT Solutions for Exercise 8.4 Class 10 Maths Chapter 8 - Introduction to Trigonometry

Q 2: Write all the other trigonometric ratios of $\mathrm{\angle }A$ in terms of $\sec A$ .

Answer:

We know that the identity ${\sin }^{2}A+{\cos }^2=1$

$$\begin{gathered} {\sin }^{2}A=1-{\cos }^2 \\ si{n}^{2}A=1-\frac{1}{{\sec }^{2}A} \end{gathered}$$

$=\frac{{\sec }^{2}A-1}{{\sec }^{2}A}$

$\sin A=\sqrt{\frac{{\sec }^{2}A-1}{{\sec }^{2}A}}$

$=\frac{\sqrt{{\sec }^{2}A-1}}{\sec A}$

$cosecA=\frac{\sec A}{\sqrt{{\sec }^{2}A-1}}$

$\tan A=\frac{\sin A}{\cos A}=\sqrt{{\sec }^{2}A-1}$

$\cot A=\frac{1}{\sqrt{{\sec }^{2}A-1}}$

Q 3: Choose the correct option. Justify your choice $(i)9{\sec }^{2}A-9{\tan }^{2}A=$

(A) 1 (B) 9 (C) 8 (D) 0

Answer:

The correct option is (B) = 9

$9{\sec }^{2}A-9{\tan }^{2}A=9({\sec }^{2}A-{\tan }^{2}A)$ .............(i)

And it is known that $se{c}^{2}A-ta{n}^{2}A=1$

Therefore, equation (i) becomes, $9\times 1=9$

Q 3: Choose the correct option. Justify your choice $(ii)(1+\tan \theta +\sec \theta )(1+\cot \theta -cosec\theta )=$

(A) 0 (B) 1 (C) 2 (D) –1

Answer:

The correct option is (C)

$(1+\tan \theta +\sec \theta )(1+\cot \theta -cosec\theta )$ .......................(i)

we can write his above equation as;

$$\begin{gathered} =(1+\sin \theta /\cos \theta +1/\cos \theta )(1+\cos \theta /\sin \theta -1/sin\theta ) \\ =\frac{(1+\sin \theta +\cos \theta )}{\cos \theta .\sin \theta }\times (\frac{(\sin \theta +\cos \theta -1}{\sin \theta .\cos \theta }) \\ =\frac{(\sin \theta +\cos \theta {)}^2-1^2}{\sin \theta .\cos \theta } \\ =\frac{{\sin }^2\theta +{\cos }^2\theta +2\sin \theta . \\ cos\theta -1}{\sin \theta .\cos \theta } \\ =2\times \frac{\sin \theta .\cos \theta }{\sin \theta .\cos \theta } \end{gathered}$$ = 2

Q 3: Choose the correct option. Justify your choice $(iii)(\sec A+\tan A)(1-\sin A)=$

$(A)\sec A$ $(B)\sin A$ $(C)cosecA$ $(D)\cos A$

Answer:

The correct option is (D)

$(\sec A+\tan A)(1-\sin A)=$

$$\begin{gathered} \Rightarrow (\frac{1}{\cos A}+\frac{\sin A}{\cos A})(1-\sin A) \\ \Rightarrow \frac{1+\sin A}{\cos A}(1-\sin A) \\ \Rightarrow \frac{1-{\sin }^{2}A}{\cos A} \\ \Rightarrow \cos A \end{gathered}$$

Q 3: Choose the correct option. Justify your choice $(iv)\frac{1+{\tan }^{2}A}{1+{\cot }^{2}A}=$

$(A){\sec }^{2}A$ $(B)-1$ $(C){\cot }^{2}A$ $(D){\tan }^{2}A$

Answer:

The correct option is (D)

$\frac{1+{\tan }^{2}A}{1+{\cot }^{2}A}$ ..........................eq (i)

The above equation can be written as;

We know that $\cot A=\frac{1}{\tan A}$

therefore,

$$\begin{gathered} \Rightarrow \frac{1+{\tan }^{2}A}{1+\frac{1}{{\tan }^2}A} \\ \Rightarrow {\tan }^{2}A\times (\frac{1+{\tan }^{2}A}{1+{\tan }^{2}A}) \\ \Rightarrow {\tan }^{2}A \end{gathered}$$

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined $(i)(\csc \theta -\cot \theta {)}^2=\frac{1-\cos \theta }{1+\cos \theta }$

Answer:

We need to prove-

$(\csc \theta -\cot \theta {)}^2=\frac{1-\cos \theta }{1+\cos \theta }$

Now, taking LHS,

$(\csc \theta -\cot \theta {)}^2=(\frac{1}{\sin \theta }-\frac{\cos \theta }{\sin \theta }{)}^2$

$$\begin{gathered} =(\frac{1-\cos \theta }{\sin \theta }{)}^2 \\ =\frac{(1-\cos \theta )(1-\cos \theta )}{{\sin }^2\theta } \end{gathered}$$

$$\begin{gathered} =\frac{(1-\cos \theta )(1-\cos \theta )}{1-{\cos }^2\theta } \\ =\frac{(1-\cos \theta )(1-\cos \theta )}{(1-\cos \theta )(1+\cos \theta )} \\ =\frac{1-\cos \theta }{1+\cos \theta } \end{gathered}$$

LHS = RHS

Hence proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined $(ii)\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}=2\sec A$

Answer:

We need to prove-

$\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}=2\sec A$

Taking LHS;

$$\begin{gathered} =\frac{{\cos }^{2}A+1+{\sin }^{2}A+2\sin A}{\cos A(1+\sin A)} \\ =\frac{2(1+\sin A)}{\cos A(1+\sin A)} \\ =2/\cos A=2\sec A \end{gathered}$$

= RHS

Hence proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined $(iii)\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\sec \theta \csc \theta$

[ Hint: Write the expression in terms of $\sin \theta$ and $\cos \theta$ ]

Answer:

We need to prove-

$\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\sec \theta cosec\theta$

Taking LHS;

$$\begin{gathered} \Rightarrow \frac{{\tan }^2\theta }{\tan \theta -1}+\frac{1}{\tan \theta (1-\tan \theta )} \\ \Rightarrow \frac{{\tan }^3\theta -{\tan }^4\theta +\tan \theta -1}{(\tan \theta -1).\tan \theta .(1-\tan \theta )} \\ \Rightarrow \frac{({\tan }^3\theta -1)(1-\tan \theta )}{\tan \theta .(\tan \theta -1)(1-\tan \theta )} \end{gathered}$$

By using the identity a ³ - b ³ =(a - b) (a ² + b ² +ab)

$$\begin{gathered} \Rightarrow \frac{(\tan \theta -1)({\tan }^2\theta +1+\tan \theta )}{\tan \theta (\tan \theta -1a)} \\ \Rightarrow \tan \theta +1+\frac{1}{\tan \theta } \\ \Rightarrow 1+\frac{1+{\tan }^2\theta }{\tan \theta } \\ \Rightarrow 1+{\sec }^2\theta \times \frac{1}{\tan \theta } \\ \Rightarrow 1+\sec \theta .\csc \theta \end{gathered}$$

= RHS

Hence proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined $(iv)\frac{1+\sec A}{\sec A}=\frac{{\sin }^{2}A}{1-\cos A}$

[ Hint : Simplify LHS and RHS separately]

Answer:

We need to prove-

$\frac{1+\sec A}{\sec A}=\frac{{\sin }^{2}A}{1-\cos A}$

Taking LHS;

$$\begin{gathered} \Rightarrow \frac{1+\sec A}{\sec A} \\ \Rightarrow (1+\frac{1}{\cos A})/\sec A \\ \Rightarrow 1+\cos A \end{gathered}$$

Taking RHS;

We know that identity $1-{\cos }^2\theta ={\sin }^2\theta$

$$\begin{gathered} \Rightarrow \frac{{\sin }^{2}A}{1-\cos A} \\ \Rightarrow \frac{1-{\cos }^{2}A}{1-\cos A} \\ \Rightarrow \frac{(1-\cos A)(1+\cos A)}{(1-\cos A)} \\ \Rightarrow 1+\cos A \end{gathered}$$

LHS = RHS

Hence proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined $(v)\frac{\cos A-\sin A+1}{\cos A+\sin A-1}=\csc A+\cot A$ , using the identity ${\csc }^{2}A=1+{\cot }^{2}A$

Answer:

We need to prove -

$\frac{\cos A-\sin A+1}{\cos A+\sin A-1}=cosecA+\cot A$

Dividing the numerator and denominator by $\sin A$ , we get;

$$\begin{gathered} =\frac{\cot A-1+\csc A}{\cot A+1-\csc A} \\ =\frac{(\cot A+\csc A)-({\csc }^{2}A-{\cot }^{2}A)}{\cot A+1-\csc A} \\ =\frac{(\csc A+\cot A)(1-\csc A+\cot A)}{\cot A+1-\csc A} \\ =\csc A+\cot A \\ =RHS \end{gathered}$$

Hence Proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined $(vi)\sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A$

Answer:

We need to prove -

$\sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A$

Taking LHS;

By rationalising the denominator, we get;

$$\begin{gathered} =\sqrt{\frac{1+\sin A}{1-\sin A}\times \frac{1+\sin A}{1+\sin A}} \\ =\sqrt{\frac{(1+\sin A{)}^2}{1-{\sin }^{2}A}} \\ =\frac{1+\sin A}{\cos A} \\ =\sec A+\tan A \\ =RHS \end{gathered}$$

Hence proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined. $(vii)\frac{\sin \theta -2{\sin }^3\theta }{2{\cos }^3\theta -\cos \theta }=\tan \theta$

Answer:

We need to prove -

$\frac{\sin \theta -2{\sin }^3\theta }{2{\cos }^3\theta -\cos \theta }=\tan \theta$

Taking LHS;

[we know the identity $\cos 2\theta =2{\cos }^2\theta -1={\cos }^2\theta -{\sin }^2\theta$ ]

$$\begin{gathered} \Rightarrow \frac{\sin \theta (1-2{\sin }^2\theta )}{\cos \theta (2{\cos }^2\theta -1)} \\ \Rightarrow \frac{\sin \theta ({\sin }^2\theta +{\cos }^2\theta -2{\sin }^{\theta })}{\cos \theta .\cos 2\theta } \\ \Rightarrow \frac{\sin \theta .\cos 2\theta }{\cos \theta .\cos 2\theta } \\ \Rightarrow \tan \theta =RHS \end{gathered}$$

Hence proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined $(viii)(\sin A+\csc A{)}^2+(\cos A+\sec A{)}^2=7+{\tan }^{2}A+{\cot }^{2}A$.

Answer:

Given the equation,

$(\sin A+\csc A{)}^2+(\cos A+\sec A{)}^2=7+{\tan }^{2}A+{\cot }^{2}A$ ..................(i)

Taking LHS;

$(\sin A+\csc A{)}^2+(\cos A+\sec A{)}^2$

$$\begin{gathered} \Rightarrow {\sin }^{2}A+{\csc }^{2}A+2+{\cos }^{2}A+{\sec }^{2}A+2 \\ \Rightarrow 1+2+2+(1+{\cot }^{2}A)+(1+{\tan }^{2}A) \end{gathered}$$

[since ${\sin }^2\theta +{\cos }^2\theta =1,{\csc }^2\theta -{\cot }^2\theta =1,{\sec }^2\theta -{\tan }^2\theta =1$ ]

$$\begin{gathered} 7+{\csc }^{2}A+{\tan }^{2}A \\ =RHS \end{gathered}$$

Hence proved

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined $(ix)(cosecA-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}$

[ Hint : Simplify LHS and RHS separately]

Answer:

We need to prove-

$(coescA-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}$

Taking LHS;

$$\begin{gathered} \Rightarrow (cosecA-\frac{1}{\csc A})(\sec A-\frac{1}{\sec A}) \\ \Rightarrow \frac{(cose{c}^2-1)}{cosecA}\times \frac{{\sec }^{2}A-1}{\sec A} \\ \Rightarrow \frac{{\cot }^{2}A}{cosecA}.\frac{{\tan }^{2}A}{\sec A} \\ \Rightarrow \sin A.\cos A \end{gathered}$$

Taking RHS;

$$\begin{gathered} \Rightarrow \frac{1}{\sin A/\cos A+\cos A/\sin A} \\ \Rightarrow \frac{\sin A.\cos A}{{\sin }^{2}A+{\cos }^{2}A} \\ \Rightarrow \sin A.\cos A \end{gathered}$$

LHS = RHS

Hence proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined $(x)(\frac{1+{\tan }^{2}A}{1+{\cot }^{2}A})=(\frac{1-\tan A}{1-\cot A}{)}^2={\tan }^{2}A$

Answer:

We need to prove,

$(\frac{1+{\tan }^{2}A}{1+{\cot }^{2}A})=(\frac{1-\tan A}{1-\cot A}{)}^2={\tan }^{2}A$

Taking LHS;

$\Rightarrow \frac{1+{\tan }^{2}A}{1+{\cot }^{2}A}=\frac{{\sec }^{2}A}{{\csc }^{2}A}={\tan }^{2}A$

Taking RHS;

$$\begin{gathered} =(\frac{1-\tan A}{1-\cot A}{)}^2 \\ =(\frac{1-\sin A/\cos A}{1-\cos A/\sin A}{)}^2 \\ =\frac{(\cos A-\sin A{)}^2({\sin }^{2}A)}{(\sin A-\cos A{)}^2({\cos }^{2}A)} \\ ={\tan }^{2}A \end{gathered}$$

LHS = RHS

Hence proved.